Question

find the value of discriminant of each of quadratic equation
1) x^2-6x+k=0
2)9x^2-1=0
please full method. ​

Answer 1

Answer:

1) x^2 - 6x + k = 0

discriminant = b^2 - 4 ac

=> (-6)^2 - 4 * 1 * k = 0

=> 36 - 4k = 0

=> -4k = -36

=> 4k = 36

= > k = 9

hence the value of k is 9.

2) 9x^2 - 1 = 0

discriminant = b^2 -4ac

=> ( 0 )^2 - 4 *9* -1 =0

=> 0 + 36 = 0

=> -36

hence the value of discriminant is negative..

so the value of x has no real root.

hope it helps

Answer 2

Step-by-step explanation:

discriminant=b^2-4ac for ax^2+bx+c=0

1. discriminant=(-6)^2-4(1)(k)=36-4k
2. discriminant=0-4(9)(-1)=36.